Math, asked by chimmi6072, 8 months ago

Arvind deposited a sum of money with a bank on 1st january,2012 at 8% simple interest per annum. He received an amount RS. 3.144 on 7th August,2012. ​

Answers

Answered by bhagyashreechowdhury
0

The formula required to solve the problem:

  • Simple Interest = \frac{P\:*\:R\:*\:T}{100}
  • Amount = Principal + Interest

Given:

Arvind deposited the sum of money with the bank on 1st January 2012  

The rate of interest, R = 8% p.a.

The amount received by Arvind on 7th August 2012 = Rs. 3144

To find:

The sum of money deposited by Arvind with the bank

Solution:

As given in the question that the sum was deposited on 1st January 2012 and the amount was received on 7th August 2012, therefore the time for which the sum was deposited, will be given by

= \frac{31\:+\:29\:+\:31\:+\:30\:+\:31\:+\:30\:+\:31\:+\:6\:}{366} \\= \frac{219}{366}

Let “P” be the sum of money deposited by Arvind.

Now, combining both the formula of simple interest and substituting the given values, we get

Amount – Principal = \frac{P\: *\:R\: *\:T}{100}

3144 - P = \frac{P\:*\:8\:*\:\frac{219}{366} }{100}

3144 - P = \frac{P\:*\:8\:*\:0.598}{100}

⇒ 314400 – 100P = 4.784P

⇒ 314400 = 100P + 4.789P

⇒ 104.784P = 314400

⇒ P = \frac{314400}{104.784}

P = Rs. 3000.45 which is approximately Rs. 3000

Thus, the money he deposited with the bank was Rs. 3000.

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